#lim_(n->oo)prod_(k=2)^n(1-1/k^2)#?

2 Answers
Jul 13, 2018

Answer:

#prod_(k=2)^oo (1-1/k^2) = 1/2#

Explanation:

Consider the sequence of integrals:

#I_m = int_0^(pi/2) sin^mx dx#

As the integrand is defined and continuous in the whole interval, these integrals exist for every #m in NN#.

Evaluate now the integral by parts:

#I_m = -int_0^(pi/2) sin^(m-1)x d(cosx)#

#I_m = -[sin^(m-1)x cosx]_0^(pi/2) + int_0^(pi/2) (m-1)sin^(m-2)xcos^2xdx#

#I_m = (m-1) int_0^(pi/2) sin^(m-2)x(1-sin^2x)dx#

#I_m = (m-1) I_(m-2) - (m-1)I_m#

and finally we have the recursive formula:

#I_m = (m-1)/m I_(m-2) #

so that:

#I_(m-2)/I_m= m/(m-1)#

and:

#lim_(m->oo) I_(m-2)/I_m= 1#

Given that:

#I_0 = int_0^(pi/2) dx = pi/2#

#I_2 = int_0^(pi/2) sin^2xdx = pi/4#

consider now for every #m# the ratio:

#I_(m-1)/I_(m+1) = I_(m-1)/I_m I_m/I_(m+1)#

and evaluate each ratio using the recursive formula:

#I_(m-1)/I_(m) = ((m-2)/(m-1) I_(m-3))/((m-1)/(m) I_(m-2)#

#I_(m-1)/I_(m)= ((m(m-2))/(m-1)^2 ) I_(m-3)/I_(m-2)#

#I_(m-1)/I_(m) = (((m-1+1)(m-1-1))/(m-1)^2 ) I_(m-3)/I_(m-2)#

#I_(m-1)/I_(m) = (((m-1)^2-1)/(m-1)^2 ) I_(m-3)/I_(m-2)#

#I_(m-1)/I_(m) = (1-1/(m-1)^2 ) I_(m-3)/I_(m-2)#

and:

#I_(m)/I_(m+1)= ((m-1)/(m) I_(m-2))/((m)/(m+1) I_(m-1)#

#I_(m)/I_(m+1)= (((m+1)(m-1))/m^2) I_(m-2)/ I_(m-1)#

#I_(m)/I_(m+1)= ((m^2-1)/m^2) I_(m-2)/ I_(m-1)#

#I_(m)/I_(m+1)= (1-1/m^2) I_(m-2)/ I_(m-1)#

Then:

#I_(m-1)/I_(m+1) = (1-1/m^2) I_(m-2)/ I_(m-1)(1-1/(m-1)^2 ) I_(m-3)/I_(m-2)#

#I_(m-1)/I_(m+1) = (1-1/m^2)(1-1/(m-1)^2 ) I_(m-3)/I_(m-1)#

and iterating:

#I_(m-1)/I_(m+1) = (1-1/m^2)(1-1/(m-1)^2 )(1-1/(m-2)^2)(1-1/(m-3)^2 ) I_(m-5)/I_(m-3)#

or:

#I_(m-1)/I_(m+1) = prod_(j=0)^p (1-1/(m-j)^2) I_(m-p-2)/I_(m-p)#

So for #p=m-2# we have:

#I_(m-1)/I_(m+1) = prod_(j=0)^(m-2) (1-1/(m-j)^2) I_(0)/I_(2)#

#I_(m-1)/I_(m+1) = prod_(j=0)^(m-2) (1-1/(m-j)^2)(pi/2)/(pi/4)#

#I_(m-1)/I_(m+1) = 2 prod_(j=0)^(m-2) (1-1/(m-j)^2)#

Change now the index to #k= m-j#:

#I_(m-1)/I_(m+1) = 2 prod_(k=2)^m (1-1/k^2)#

so:

#prod_(k=2)^m (1-1/k^2) = 1/2 I_(m-1)/I_(m+1)#

and passing to the limit:

#prod_(k=2)^oo (1-1/k^2) = 1/2 lim_(m->oo) I_(m-1)/I_(m+1) = 1/2#

Jul 14, 2018

#lim_(n->oo)prod_(k=2)^n(1-1/k^2)#

#= lim_(n->oo)prod_(k=2)^n ((k-1)(k+1))/k^2#

#= lim_(n->oo) underbrace( ( (1)(3))/2^2)_(k = 2) * underbrace( ((2 )(4))/3^2)_(k = 3) * underbrace( ((3)(5))/4^2)_(k = 4) * underbrace( ((4)(6))/5^2)_(k = 5) ... * underbrace( ((n-3)(n-1))/(n-2)^2)_(k = n-2) * underbrace( ((n-2)(n))/(n-1)^2)_(k = n-1) * underbrace( ((n-1)(n+1))/n^2)_(k = n)#

#= lim_(n->oo) underbrace( ( (1))/2 )_(k = 2) * underbrace( ((1 )(1))/1^2)_(k = 3) * underbrace( ((1)(1))/1^2)_(k = 4) * underbrace( underbrace( ((1)(6))/5 )_(k = 5) * ... * underbrace( ((n-3)(1))/(n-2) )_(k = n-2) )\_(" Not fully factored ") * underbrace( ((1)(1))/(1)^2)_(k = n-1) * underbrace( ((1)(n+1))/n )_(k = n)#

The pattern : every term, in conjunction with the immediately preceding and succeeding terms, reduces to #1#. This is not true for the first and last term which are as shown.

# = lim_(n->oo) 1/2 * 1 * 1 * ... * 1 * (n+1)/ n #

# = 1/2 lim_(n->oo) 1+1/n = 1/2#